1
The
Global
Weak Solutions of the Compressible Euler
Equation with
Spherical Symmetry
Tetu Makino*, Kiyoshi Mizohata** and Seiji Ukai**
(December, 1991)
Department ofLiberal Arts
Osaka Sangyo University* and
Department of Information Sciences
Tokyo Institute of Technology**
1 Introduction
The compressible Euler equation for an isentropic
gas
in $R^{n}$ is given by$\rho_{t}+\nabla\cdot(\rho\tilde{u})=0$ ,
(1.1)
$(\rho\vec{u})_{t}+\nabla\cdot(\rho\vec{u}\otimes\vec{u}+p)=0$
,
with the equation ofstate
(1.2) $p=a^{2}\rho^{\gamma}$ ,
where density$\rho$
,
velocity$\vec{u}$and pressure
$p$ are functionsof $x\in R^{n}$ and $t\geq 0$,
while $a>0$ and $\gamma\geq 1$ are given constants.
Forone dimensional case $(n=1)$
,
the Cauchy problem for (1.1) with (1.2)has been studied by many authors. Nishida [10] established the existence of global weak solutions, for the first time, for the case $\gamma=1$ with arbitrary
initial data, and Nishida and Smoller [11] for $\gamma\geq 1$ but with small initial
data, both
using Glimm’s
method. DiPerna [3] extended the latter result tothe case of large initial data, using the theory of compensated compactness under the restriction $\gamma=1+2/(2m+1),$ $m\geq 2$ integers. Ding et al
[1], [2] removed this restriction and established the existence ofglobal weak
solutions for $1<\gamma\leq 5/3$
.
数理解析研究所講究録 第 785 巻 1992 年 1-28
On the other hand, littleis known for the case $n\geq 2$
.
No global solutionshave been known to exist, but only local classical solutions ([5], [6], [8] and
[9]).
In this paper, we will present global weak solutions first
for
the case$n\geq 2$
.
We will do this, however, only forthe
case of spherically symmetrywith $\gamma=1$
.
As will be seen below, our proof does not work without theserestrictions.
Thus, we look for solutions of the form
(1.3) $\rho=\rho(t, |x|))\tilde{u}=\frac{x}{|x|}\cdot u(t, |x|)$
.
Then, denoting $r=|x|,$ $(1.1)$ becomes(1.4) $\rho_{t}\rho+_{t}\frac{1}{r^{n-1}+u}(;^{n-}1^{\rho_{P}u_{f}})_{f}==00$ ,
This equation has a singularity at $r=0$
.
To avoid the difficulty caused bythis singularity, we simply deal with the boundary value problem for (1.4)
in the domain $1\leq r<\infty$ (the exterior of a sphere) with the boundary
condition $u(t, 1)=0$, which is identical, under the assumption (1.3), to the usual boundary condition $\tilde{n}\cdot\tilde{u}=0$ for (1.1) where $\tilde{n}$ is the unit normal to the boundary.
Put $\tilde{\rho}=r^{\mathfrak{n}-1}\rho$
.
Then we get from (1.4)$\tilde{\rho}_{\ell}+(\tilde{\rho}u)_{r}=0$ , (1.5)
$u_{\ell}+u u_{f}+\frac{a^{2}\gamma\tilde{\rho}_{r}}{\tilde{\rho}^{2-\gamma_{\Gamma}\langle n-1)\langle\gamma-1)}}=\frac{a^{2}\gamma(n-1)\tilde{\rho}^{\gamma-1}}{r^{n}\cdot r^{(\mathfrak{n}-1)(\gamma-2)}}$
Introduce the Lagrangean mass coordinates
(1.6) $\tau=t$
,
$\xi=l^{r}\tilde{\rho}(t,r)$ dr.Then $\xi>0$ as long as $\tilde{\rho}>0$ for $r>1$, and (1.5) is
reformulated
as $\tilde{\rho}_{\tau}+\tilde{\rho}^{2}u_{\xi}=0$ ,(1.7)
3
Put $v=1/\tilde{\rho}$ and note that the inverse transformation to (1.6) is
given
by
(1.8) $t=\tau,$ $r=1+ \int_{0}^{\epsilon}v(\zeta,t)d\zeta$
.
Then after changing $\tau$ to $t$ and $\xi$ to
$x,$ $(1.7)$ is written as
$v_{t}-u_{x}=0$ ,
(1.9)
$u_{t}+( \frac{a^{2}}{v^{\gamma}})_{x}$ $\frac{1}{r^{(n-1)(\gamma-1)}}=\frac{a^{2}\gamma(n-1)v^{1-\gamma}}{r^{n}\cdot r^{(n-1)(\gamma-2)}}$
where $r$ is now defined by $r=1+ \int_{0}^{x}v(t, \zeta)d\zeta$
.
Now we restrict ourseves to the case $\gamma=1$
.
Then (1.7) becomes$v_{t}-u_{x}=0$
,
(1.10)
$u_{t}+( \frac{a^{2}}{v})_{x}=\frac{K}{1+\int_{0}^{x}v(t,\zeta)d\zeta}$
where $K=a^{2}(n-1)$
.
Let us consider the initial boundary value problrm for (1.10) in
$t\geq 0,$ $x\geq 0$ with the following boundary and initial conditions.
(1.11) $u(O, x)=u_{0}(x)$ , $v(O, x)=v_{0}(x)$ ,
for
$x>0$,(1.12) $u(t, 0)=0$ ,
for
$t>0$.Let $BV(R_{+})$ denote the space of functions of bounded variation on
$R_{+}=(0, \infty)$
.
Our main result is as follows.Theorem (Main Result) Suppose that $u_{0}(x),$ $v_{0}(x)\in BV(R_{+})$
,
and that$v_{0}(x)\geq\delta_{0}>0$
for
all $x>0$ with some positive constant$\delta_{0}$.
Then (1.10),(1.11) and (1.12) have a global weak solution which belongs to the class
$u,$ $v\in L^{\infty}(0, T;BV(R_{+}))\cap Lip([0, T];L_{loc}^{1}(R_{+}))$
for
any$T>0$.
The definition of the weak solution will be given in section 4. This the-orem can be proved by
following
Nishida’s argument [10] basedon Glimm’s
method. Indeed this can be seen from thefollowing twosimple observations.
First, the homogeneous equation corresponding to (1.10),
$v_{t}-u_{x}=0$ ,
(1.13)
$u_{t}+( \frac{a^{2}}{v})_{x}=0$ ,
is just the same equation as solved by Nishida [10] using Glimm’s method
both on
the
Cauchy problem and the initial boundary value problem. Notethat if $\gamma>1$, the homogeneous equation for (1.9) has a variable coefficient
and hence does not coincide with the one dimensional Euler equation.
The second observation is that, as long as $v\geq 0$, the right hand side of
(1.10),
(1.14) $\frac{K}{1+\int_{0}^{x}v(t,()d\zeta}$
is monotone decreasing in $x$ and has an a priori estimate
(1.15) T.V. $( \frac{K}{1+\int_{0}^{x}v(t,\zeta)d\zeta})\leq K$,
independent of $v$
.
The one dimensional inhomogeneous Euler equation hasbeen studied in [12]. However, the conditions imposed therein on the
inho-mogeneous
term are not applicable to our (1.14).These observations allow us to use Nishida’s argument [10] to construct global weak solutions to (1.10), (1.11) and (1.12). More precisely, we will
first construct, in section 2, approximate solutions of the form
{solution
of
Riemann problemfor
(1.13)} $+${nonhomogeneous
term}
$\cross t$.
This is the main idea of [12]. Then in section 3, we will estimate the total variation of the approximate solutions. Thanks to (1.15), this can be done with a slight modification of Nishida’s argument [10]. In section 4, we will show that there exists a subsequence of approximate solutions which
con-vereges
strongly in $L_{loc}^{1}$ for any finite time interval. Finally, for the sake ofcompleteness, we give in Appendix a detailed proof of two lemmas used in section
3.
These lemmas are due to Nishida [10], but their proofs are not found in the literature.5
2 The Difference Scheme
Toconstruct the approximate solutions, we shall usethedifferencescheme developed in [10]. For $l,$$h>0$
,
define$Y=$ $\{(n , m); n=1,2,3, \cdots, m=1,3,5, \cdots\}$ ,
(2.1)
$A= \prod_{(m,n)\in Y}[\{nh\}\cross((m-1)l, (m+1)l)]$
,
where $l/h$ will be determined later. Choose a point $\{a_{nm}\}\in A$ randomly,
and write $a_{nm}=(nh, c_{nm})$
.
For $n=0$, we put $c_{Om}=ml$.
We denoteapproximate solutions by $u^{l}$ and $v^{l}$
.
Mesh lengths $l$ and $h$ are chosen so that$l/h>a/( \inf v^{t})$, for anygiven $T>0$
.
We shall show later that there exists a$\delta>0$ such thatinf
$v^{l}\geq\delta>0$.
For $0\leq t<h,$ $ml\leq x<(m+2)l,$ $m$ : odd, we define
$u^{l}(t,x)=u_{0}^{l}(t, x)+U^{l}(t,x)t$,
(2.2)
$v^{l}(t,x)=v_{0}^{l}(t,x)$,
where $u_{0}^{l}$ and $v_{0}^{l}$ are the solutions of
$v_{t}-u_{x}=0$
,
(2.3)
$u_{\ell}+( \frac{a^{2}}{v})_{x}=0$,
with initial data
$u_{0}^{l}(0, x)=$ (2.4) $\{u_{0}^{0}((m+2)l)u(ml)$, $X>(mX<(m:_{1)l}^{1)l},$’ $v_{0}^{l}(0, x)=\{v_{0}^{0}((m+2)l)v(ml)$
,
$X>(mX<(mI_{1)l’}^{1)l}$ , and (2.5) $U^{l}(t,x)= \frac{K}{1+\Sigma^{\frac{m+1}{j=^{2}1}}v_{0}((2j-1)l)\cdot 2l}$For $0\leq t<h,$ $0\leq x<l$, we define $u^{l}$ and $v^{l}$ by (2.2)
where $u_{0}^{l}$ and $v_{0}^{l}$
are the solutions of (2.3) with initial boundary data
(2.7) $u(t, 0)=0,$ $t>0$, and
(2.8) $U^{l}(t,x)=K$
.
Suppose that $u^{l}$ and $v^{l}$ are defined for$0\leq t<nh$
.
For$nh\leq t<(n+1)h$,
$ml\leq x<(m+2)l,$ $m$ : odd, we define
$u^{l}(t,x)=u_{0}^{l}(t,x)+U^{l}(t,x)$
.
(t–nh),(2.9)
$v^{l}(t,x)=v_{0}^{l}(t,x)$
,
where $u_{0}^{l}$ and $v_{0}^{l}$ are the solutions of (2.3) with initial data $(t=nh)$
$u_{0}^{l}(nh, x)=$ (2.10) $\{\begin{array}{l}u^{l}(nh-0,c_{nm})u^{l}(nh-0,c_{nm+2})\end{array}$ $x>(m+1)lx<(m+1)l$, $v_{0}^{l}(nh, x)=\{v^{l}(nh-0,c_{nm}^{nm})v_{l}(nh-0,c)_{+2}$ $x>(mx<(m\ddagger^{1)l}1)l$
,
and (2.11) $U^{l}(t,x)= \frac{K}{1+\Sigma^{\frac{m+1}{j=^{2}1}}v^{l}(nh-0,c_{n2j-1})\cdot 2l}$For $nh\leq t<(n+1)h,$ $0\leq x<l$
,
we define $u^{l}$ and $v^{l}$ as(2.9) where $u_{0}^{l}$
and $v_{0}^{l}$ are the solutions of (2.3) with initial $(t=nh)$ boundary data
(2.12)$u_{0}^{l}(nh, x)=u^{l}(nh-0, c_{n1}),$ $v_{0}^{l}(nh, x)=v^{l}(nh-0, c_{n1}),$ $x>0$,
(2.13) $u(t, O)=0,$ $t>nh$,
and $U^{l}(t, x)$ is as (2.8).
3 Bounds for Approximate Solutions
System (1.6) is hyperbolic provided $v>0$, with the characteristic roots
and Riemann invariants given by
$\lambda=-\underline{a}$
$r=u+alogv$
,(3.1)
a’
7
It is well-known, [10], that all shock wave curves in the (r,s)-plane have the
same figure. (See Figure 1.) Tlie l-shock wave curve $S_{1}$, starting from
$(r_{0}, s_{0})$ can be expressed in the form
(3.2) $s-s_{0}=f(r-r_{0})$ for $r\leq r_{0}$,
and the 2-shock wave curve $S_{2}$ can also be expressed in the form
(3.3) $r-r_{0}=f(s-s_{0})$ for $s\leq s_{0}$,
where
The l-rarefaction wave curve $R_{1}$ can be expressed in the form
(3.4) $s-s_{0}=0$ for $r\geq r_{0}$,
and the corresponding expression for the 2-rarefaction wave curve $R_{2}$ is
(3.5) $r-r_{0}=0$ for $s\geq s_{0}$
.
Now we must prepare some lemmas to estimate Riemann invariants.
First, let us consider (2.3) with following initial data
(3.6) $u_{0}(x)=\{\begin{array}{l}u_{l}u_{r}\end{array}$ $v_{0}(x)=\{\begin{array}{l}v_{l},x<0v_{f},x>0\end{array}$
Lemma 3.1 Let $u$ and $v$ are the solutions
of
(2.3) and (3.6). Then,(3.7) $\{r(t,x)\equiv r(u(t,x),v(t,x))\geq ro\equiv\min(r(u,v),r(u,v_{v_{l}^{l}})_{)})_{)}s(t,x)\equiv s(u(t,x),v(t,x))\leq so\equiv\max(s(u^{f_{f}},v^{r_{f}}),s(u_{l}^{l},$
Next consider (2.3) in $t\geq 0,$ $x\geq 0$ with following initial and boundary
conditions
(3.8) $u(O, x)=u_{0}^{+}$ , $v(O, x)=v_{0}^{+}$
,
for
$x>0$,
(3.9) $u(t, 0)=0$,
for
$t>0$.
Lemma 3.2 Let $u$ and $v$ are the solutions
of
(2.3), (3.8) and (3.9). Then,(3.10) $\{\begin{array}{l}r(t,x)\equiv r(u(t,x),s(t,x))\geq r(u_{0}^{+},v_{0}^{+})s(t,x)\equiv s(u(t,x),s(t,x))\leq\max(-r(u_{0}^{+},v_{o}^{+}),s(u_{o}^{+},v_{0}^{+}))\end{array}$
The above two lemmas were proved in [10]. Using thesetwo lemmas, we can
get the following lemma. Lemma 3.3 Let $u^{l}$ and $v^{l}$
be the approximate solutions
defined
in section 2and put $r_{0}= \min r(u_{0}(x), v_{0}(x))$ and $s_{0}= \max s(u_{0}(x),v_{0}(x))$
.
Then,for
$0<t<T$,
9
Let us consider Riemann problem (2.3) and (3.6). Denote by $\Delta r$
(resp $\Delta s$) the absolute value of the variation ofthe Riemann invariant
$r$
(resp s) in the first (resp second) schock wave.
Definition 3.4 We denote
$P(u_{l}, v_{l}, u_{f}, v_{f})=\triangle r+\Delta s$
.
Then we have the following lemma.Lemma 3.5
(3.12) $P(u_{1},v_{1},u_{3}, v_{3})\leq P(u_{1}, v_{1}, u_{2}, v_{2})+P(u_{2}, v_{2}, u_{3}, v_{3})$,
where $u_{1},$ $u_{2}$ and $u_{3}$ are arbitrary constants and $v_{1},$ $v_{2}$ and $v_{3}$ are arbitrary
positive costants.
We shall prove Lemma 3.5 in the Appendix A.
Denote by $i_{0}^{n\pm}$ the straight line segmentsjoining the points $(0, (n \pm\frac{1}{2})h)$
and $a_{1n}$
.
Let $F(i_{0}^{n\pm})$ be the absolute value of the variation of the Riemanninvariants for all shocks on $i_{0}^{n\pm}$
.
Then we also have the following Lemma.Lemma 3.6
(3.13) $F(i_{0}^{n+})\leq F(i_{0}^{n-})$
.
This lemma 3.6 will be proved in the Appendix B.
We denote
$Z_{1}=\{l-O, l+O, 3l-0, \cdots , (2m-1)l-0, (2m-1)l+0, \cdots\}$,
$Z_{2}=\{2l, 4l, 6l\cdots 2ml, \cdots\}$
.
Let $Z_{t^{n)}}=Z_{1}\cup Z_{2}\cup\{c_{nm}\}$ and line up the elements $z_{n,i}$ of $Z_{(n)}$ so that
$z_{n,i}\leq z_{n,i+1}$
.
(We regard$(2m-1)l-0<(2m-1)l+0$
for $m$ : integer. ) Let$F(nh- O,u^{l}, v^{l})=\frac{1}{2}F(i_{0}^{n-})$
$F(nh+0, u^{l}, v^{l})= \frac{1}{2}F(i_{0}^{n+})+\sum_{m:odd}P(u^{l}(a_{nm}), v^{l}(a_{nm}),$ $u^{l}(a_{nm+2}),$$v^{l}(a_{mm+2}))$
.
Using Lemma 3.5 and Lemma 3.6, we get
(3.14) $F((n+1)h+0, u^{l}, v^{l})\leq F((n+1)h-0,u^{l}, v^{l})$
.
The followingequality is obvious from the definition of $F,$ $u^{l}$ and $v^{l}$
.
(3.15) $F((n+1)h-0, u_{0}^{l}, v_{0}^{l})=F(nh+0, u^{l}, v^{l})$
.
We also get
$F((n+1)h-0, u^{1}, v^{l})=F((n+1)h-0, u_{0}^{l}, v_{0}^{l})$
$+ \sum_{m:odd}P$(
$u^{l}$($(n+1)h-0$,ml–O),$v^{l}(n+1)h-0$,ml–O),
$u^{l}(n+1)h-0,ml+0),v^{l}((n+1)h-0, ml+0))$.
Lemma 3.7
$P(u^{l}$($(n+1)h-0$
,
ml–O), $v^{l}$( $(n+1)h-0$,
ml–O),
(3.16) $u^{l}((n+1)h-0,ml+0),v^{l}((n+1)h-0,ml+0)$
$\leq 2h\{U^{l}(nh, (m-1)l)-U^{l}(nh, (m+1)l)\},$ $m$ : odd.
Proof.
From the definition,$u^{l}$( $(n+1)h-0$,ml-O) $=u_{0}^{l}(nh, ml)+U^{l}(nh, (m-1)l)\cdot h$,
$u^{l}((n+1)h-0, ml+0)=u_{0}^{l}(nh, ml)+U^{l}(nh, (m+1)l)\cdot h$
,
$v^{l}$($(n+1)h-0$,ml–O) $=v^{l}((n+1)h-0, ml+0)=v_{0}^{l}(nh, ml)$
.
Therefore we get
$r^{l}$(
$(n+1)h-0$
,
ml–O) $-r^{l}((n+1)h-0,ml+0)$(3.17) $=s^{l}$($(n+1)h-0$
,
ml–O) $-s^{l}((n+1)h-0, ml+0)l$$=h\cross\{U^{l}(nh, (m-1)l)-U^{l}(nh, (m+1)l)\}\geq 0$
Thus thefollowing inequality holds.
$\int[$
From (3.18), we get (3.16). $\square$
Using Lemma 3.7, we get
$F((n+1)h-0, u^{l},v^{l})-F((n+1)h-0, u_{o}^{l}, v_{0}^{l})$
(3.19)
$\leq 2h\sum_{m:odd}\{U^{l}(nh, (m-1)l)-U^{l}(nh, (m+1)l)\}\leq 2Kh$
From (3.14), (3.15) and (3.19), we get
(3.20) $F((n+1)h+0, u^{l}, v^{l})\leq F(nh+0, u^{l}, v^{l})+2Kh$
Thus we obtain the following lemma. Lemma 3.8
(3.21) $F(nh+0, u^{l}, v^{l})\leq F(+0, u^{l}, v^{l})+2KT\equiv F_{0}+2KT$
Denote by $G(\tau)$ the absolute value of the sum ofnegative variation of$r^{l}$ and
$s^{l}$
for $t=\tau$
.
Then for $nh\leq\tau<(n+1)h$, we get$($
3.22
$)^{G(\tau)\leq G(nh)+2h\sum_{m:odd}\{U^{l}(nh,(m-1)l)-U^{l}(nh,(m+1)l)\}}$
$\leq G(nh)+2Kh$.
Lemma 3.9
(3.23) $G(nh)\leq 2F(nh+0,u^{l}, v^{l})$
.
Proof.
Denote by $\delta s$ (resp $\delta r$ ) the absolute value ofthe Riemann invariant$s$ (resp r) in the first (resp second) shock wave. By (3.2) and (3.3),
$\triangle r+\delta s<2\triangle r$ on the first shock and $\delta r+\triangle s<2\triangle s$ on the second shock.
So from (3.17), (3.18) and above arguements, we get (3.23). $\square$
From (3.23), (3.24) and (3.25), for any $\tau(nh\leq\tau<(n+1)h)$,
$G(\tau)\leq G(nh)+2Kh\leq 2F(nh+0, u^{l}, v^{l})+2Kh$
(3.24)
$\leq 2F_{0}+6KT\equiv M_{1}$
.
Now we can establish a priori estimates of $u^{l}$ and $v^{l}$
.
Denote by T.V.$u$
Theorem 3.10 For any $T>0$, the variation
of
$u^{l}$ and $v^{l}$ is bounded
uni-formly
for
$h$ and $\{a_{mn}\}$.
Their upper bound and lower bound, especially thepositive lower bound
of
$v^{l}$, are also uniformly bounded.Proof.
Denote by $T.V^{+}.u$ (resp $T.V^{-}.u$ ) the absolute value of the positive(resp negative) variation of$u$
.
Put $f^{l}\equiv 2u^{l}=r^{l}+s^{l}$.
Then $0\leq f^{l}(t, 0)\leq$$Kh$
.
Without loss of generality, we assume that $u_{0}(x)$ and $v_{0}(x)$ are constant outside a bounded interval. Let(3.25) $f^{l}(t, \infty)=r^{l}(t, \infty)+s^{l}(t, \infty)\equiv M_{2}$.
Then from the definition,
$f^{l}(t, 0)+T.V^{+}.f^{\int}$ –T.$V^{-}.f^{l}=f^{l}(t, \infty)$
.
Since T.$V^{-}.f^{l}(t, \cdot)\leq G(t)$ for any $t,$ $(3.24)$ yields$T.V^{+}.f^{l}=f^{l}(t, \infty)+T.V^{-}.f^{l}-f^{l}(t, 0)\leq M_{1}+M_{2}$
.
Thus we get
(3.26) T.V.$f^{l}=T.V2u^{l}\leq 2M_{1}+M_{2}$
.
From (3.26), we get
$|f^{l}|\leq Kh+2M_{1}+M_{2}\leq KT+2M_{1}+M_{2}\equiv 2M_{3}$
.
Therefore we get
(3.27) $|u_{l}|\leq M_{3}$.
Using Lemma 3.2, we get
$2alogv^{l}=r^{l}-s^{l} \geq r_{0}-(\max(-r_{0}, s_{0})+KT)$
.
Thus we get
(3.28) $v^{l} \geq\exp\frac{r_{0}-(\max(-r_{0},s_{0})+KT)}{2a}\equiv\frac{1}{M_{5}}$ From the definition,
13
Using Lemma 3.3 and (3.24),
(3.29) $T.V^{+}.r^{l}=-r^{1}(0)+T.V^{-}.r^{l}+r(t, \infty)\leq-r_{0}+M_{1}+r(t, \infty)$
.
In view of (3.27) and (3.29), there exists a positive constant $M_{6}$ such that
(3.30) $v^{l}\leq M_{6}$
$\square$
Theorem 3.11 For any interval $[x_{1}, x_{2}]\subset[O.\infty$), we get
(3.31) $\int_{x_{1}}^{x_{2}}|u^{l}(t_{2}, x)-u^{l}(t_{1}, x)|+|v^{l}(t_{2}, x)-v^{l}(t_{1}, x)|dx$
$\leq M\cdot(|t_{2}-t_{1}|+h)$, $0\leq t_{1},t_{2}<T$,
where $M$depends on $T,$ $x_{1}$ and $x_{2}$, but not on $l$ and $h$
.
Proof.
Without loss ofgenerality, we assume that$nh\leq t_{1}<(n+1)h<\cdots<(n+k)h\leq t_{2}<(n+k+1)h$
.
Let
$\int_{x_{1}}^{x_{2}}|u^{l}(t_{2},x)-u^{l}(t_{1}, x)|dx$
$\leq I_{1}+I_{2}+\int_{x_{1}}^{x_{2}}|u^{l}(t_{2}, x)-u^{l}((n+k)h+0,x)|+|u^{l}(t_{1}, x)-u^{l}((n+1)h-0,x)|dx$
where
$I_{1}= \int_{x_{1}}^{x_{2}}\sum_{1=1}^{k}|u^{l}((n+i)h+O, x)-u^{l}((n+i)h-O,x)|dx$
$I_{2}= \int_{x_{1}}^{x_{2}}\sum_{1=1}^{k-1}|u^{l}((n+i+1)h-O, x)-u^{l}((n+i)h+O, x)|dx$
and
Denote by $1_{1^{\alpha,\beta]}}$ the characteristic function of the interval $[\alpha,\beta]$
.
We regard $T.V.-l<x<l=T.V_{0<x<l}$
.
Then,$I_{1}$
$\leq\sum_{1=0}^{k+1}\sum_{m:integer}\int_{x_{1}}^{x_{2}}T.V_{2ml<x<(2m+2)l}u^{l}((n+i)h-0,x)\cdot 1_{[2ml,(2m+2)l]}dx$ ,
$\leq([\frac{t_{2}-t_{1}}{h}]+2)\cdot(\sup_{0\leq t\leq T}T.Vu^{l}(t, \cdot))\cdot 2l$
.
$I_{2}$
$\leq\sum_{i_{\overline{k}}^{-0}}^{k}\sum_{m}\int_{x_{1}^{x_{2}}}(T.V_{(2m-1)l<x<(2m+1)l}u_{0}^{l}((n+i+1)h-0, x)\cdot 1_{[(2m-1)l,(2m+1)}\eta+Kh)dx$,
$\leq\sum_{:=0}2l$
.
T.$V.u_{0}^{l}((n+i+1)h-0, \cdot)+K(x_{2}-x_{1})h$,$\leq([\frac{t_{2}-t_{1}}{h}]+1)\cdot(2l\sup_{0\leq t\leq T}T.V.u_{0}^{l}(t, \cdot)+K(x_{2}-x_{1})h)$
.
The remaining terms can be evaluated similarly. For
$\int_{x_{1}}^{x_{2}}|v^{l}(t_{2}, x)-v^{l}(t_{1}, x)|dx$,
we also have a similar estimate. Combining these results gives (3.31). $\square$
4 Convergence of The Approximate Solution
Let $h_{n}=T/n$ and $h_{n}/l_{n}=\tilde{\delta}<\delta\equiv 1/M_{5}$
.
Consider the sequence$(u^{l_{n}}, v^{l_{n}})(n=1,2, \cdots)$
.
Then from Theorem 3.9 and Theorem 3.10, there exists a subsequence which converges in $L_{loc}^{1}$ to functions (u,v) uniformly for$t\in[0,T]$
.
Now we shall prove that u(x,t) and v(x,t) are the weak solutionsof initial boundary value problem (1.6), (1.7) and (1.8) provided $\{a_{nm}\}$ is
suitably chosen, namely, they satisfy the integral identity
(4.1)
$\int_{0}^{T}\int_{0}^{\infty}u\phi_{t}+(\frac{a^{2}}{v})\phi_{x}+\frac{K}{1+\int_{0}^{x}v(t,\zeta)d\zeta}\cdot\phi dxdt$
$+ \int_{0}^{\infty}u_{0}(x)\phi(0,x)dx=0$,
15
for any smooth functions $\phi$ and $\psi$ with compact support in the region
$\{(t, x) : 0\leq t<T, 0\leq x<\infty\}$ and $\phi(t,0)=0$
.
Now we know that $u_{0}^{l}$ and$v_{0}^{l}$ are weak solutions in each time strip $nh\leq t<(n+1)h$ so that for each
test function $\phi$ satisfying $\phi(t, 0)=0$,
$\int_{nh}^{(n+1)h}\int_{0}^{\infty}u^{l}\phi_{t}+(\frac{a^{2}}{v^{l}})\phi_{x}+U^{l}(t,x)\cdot\phi dxdt$
(43) $+ \int^{\infty}u^{l}(nh+O, x)\phi(nh, x)$
$- \int_{0}^{\infty}u^{\int}((n+1)h-0,x)\phi((n+1)h, x)dx=0$
Ifwe sum this over $n$, weget
$\int_{0}^{T}\int_{0}^{\infty}u^{l}\phi_{t}+(\frac{a^{2}}{v^{l}})\phi_{x}+U^{l}(t,x)\cdot\phi dxdt+\int_{0}^{\infty}u^{l}(0, x)\phi(0,x)$
(4.4)
$=- \sum_{k=1}^{N}\int_{0}^{\infty}\{u^{l}(kh+O, x)-u^{l}(kh-0, x)\}\cdot\phi(kh, x)dx$
where $N=T/h$. When $Narrow\infty$, the right-hand side of the above equality
tends to $0$ for almost every $\{a_{nm}\}\in A$ (see [4]). It is immediate to see that
$\int_{0}^{\infty}u^{l}(0, x)\phi(0,x)dxarrow\int_{0}^{\infty}u_{0}(x)\phi(0, x)dx$ $(Narrow\infty)$
.
Lemma 4.1
(4.5) $U^{l}(t,x) arrow\frac{K}{1+\int_{0}^{x}v(t,\zeta)d\zeta}$ $(Narrow\infty)$
.
locally uniformly
for
$t$ and $x$.Proof.
Let $nh\leq t<(n+1)h,$ $x\in((m-1)l, (m+1)l),$ $m$ : odd. Then(4.6) $| \int_{0}^{x}v^{l}(nh, \zeta)d\zeta-\frac{m+1}{\sum_{j=1}^{2}}v^{l}(nh, c_{2j-1n})|\leq\Vert v^{l}||_{\infty}\cdot l$.
On
the other handlocally uniformly for $t$ and $x$
.
We get
$| \int_{0}^{x}v^{l}(t, \zeta)d\zeta-\int_{0}^{x}v^{l}(nh, \zeta)d\zeta|$
(4.8) $\leq\int_{0^{\sum_{m:odd}T.V_{(m-1)l<(<(m+1)\iota^{v^{l}(nh,\cdot)\cdot 1_{1(m-1)l,\langle m+1)T}d\zeta}}}}^{x}$
.
$\leq\sup_{0\leq t\leq T}T.Vv^{\int}\cdot 2l$.
From (4.6), (4.7) and (4.8), we get (4.5). $\square$
For each test function $\psi,$ $v^{l}$ also satisfies,
$\int_{0}^{T}\int_{0}^{\infty}(v^{l}\psi_{t}-u^{l}\psi_{x})dxdt+\int_{0}^{\infty}v^{l}(0, x)\psi(O, x)dx$
(49) $=- \sum_{k=1}^{N}\int_{0}^{\infty}\{v^{l}(kh+O, x)-v^{l}(kh-0, x)\}\cdot\psi(kl, x)dx$
$-I_{1}-I_{2}$
.
where
$I_{1}= \sum_{n=0}^{N-1}\int_{nh}^{(n+1)h}U^{l}(t, O)(t-nh)\psi(t, O)dt$
and
$I_{2}= \sum_{n=0}^{N-1}\sum_{m:odd}\int_{nh}^{(n+1)h}\{U^{l}(t,ml+O)-U^{l}(t, ml-0)\}(t-nh)\psi(t,ml)dt$
.
The first term of the the right-hand side of equality (4.9) tends to $0$ for
almost every $\{a_{nm}\}\in A$ (see [4]). It is also immediate to see that
$\int_{0}^{\infty}v^{l}(O, x)\psi(O,x)dxarrow\int_{0}^{\infty}v_{0}(x)\psi(0, x)dx$ $(Narrow\infty)$
.
We shall show that $I_{1},$ $I_{2}arrow 0$ as $Narrow\infty$
.
$I_{1} \leq||\psi\Vert_{\infty}\sum_{n=0}^{N-1}\int_{nh}^{(n+1)h}U^{l}(t, O)(t-nh)dt$
(4.10)
$\leq||\psi\Vert_{\infty}\sum_{n=0}^{N-1}\int_{nh}^{(n+1)h}K(t-nh)dt$
$t7$
$\sum_{m:odd}\int_{nh}^{\langle n+1)h}\{U^{l}(t,ml+0)-U^{l}(t,ml-0)\}(t-nh)\psi(t,ml)dt\leq K\Vert\psi||_{\infty}h^{2}$.
Thus we get
(4.11) $I_{2} \leq\Vert\psi\Vert_{\infty}\sum_{n=0}^{N-1}Kh^{2}\leq K\Vert\psi\Vert_{\infty}hT$
From above arguments, we can conclude that $u$ and $v$ satisfy (4.1) and
(4.2). Thus we obtain our main result.
Theorem 4.2 (Main Result) Suppose that $u_{0}(x),$ $v_{0}(x)\in BV(R_{+})$ , and
that$v_{0}(x)\geq\delta_{0}>0$
for
all $x>0$ with somepositive constant$\delta_{0}$.
Then (1.10),(1.11) and (1.12) have a globalweak solution which belongs to the class
$u,$ $v\in L^{\infty}(O, T;BV(R_{+}))\cap Lip([0, T];L_{loc}^{1}(R_{+}))$
Appendix
A Proofof Lemma 3.5
Let
$g(x)=-f(-x)$
, and put$P(u_{1}, v_{1}, u_{2}, v_{2})=\Delta r_{1}+\Delta s_{1}$
$P(u_{2}, v_{2}, u_{3}, v_{3})=\triangle r_{2}+\Delta s_{2}$
$P(u_{1}, v_{1}, u_{3},v_{3})=\Delta r_{3}+\Delta s_{3}$
Then it is obvious that
$\Delta r_{3}+g(\Delta s_{3})+\Delta s_{3}+g(\Delta r_{3})$
$\leq\triangle r_{1}+\Delta r_{2}+\triangle s_{1}+\triangle s_{2}++g(\Delta r_{1})+g(\triangle r_{2})+g(\triangle s_{1})+g(\Delta s_{2})$
We notice that $f^{u}\leq 0$ and hence
$\leq\Delta r_{1}+\Delta r_{2}+\Delta s_{1}+\triangle s_{2}+g(\Delta r_{1}+\triangle r_{2})+g(\triangle s_{1}+\triangle s_{2})$
.
Let $x+g(x)=h(x),$ $\triangle r_{3}=p’,$ $\triangle s_{3}=q’,$ $\triangle r_{1}+\Delta r_{2}=p$ and $\Delta s_{1}+\Delta s_{2}=q$
.
Then(A.1) $h(p’)+h(q’)\leq h(p)+h(q)$
.
Put
$K=h(p’)+h(q’)$.
We shall estimate $p+q$ from below under therestriction (A.1). To do this, as $h$ is monotone increasing function, we must
estimate$p+q$ from below under the restriction
(A.2)
$h(p)+h(q)=K$
.We do this by using Lagrange’s method ofindeterminate coefficients. Put $G(p, q, \lambda)=p+q+\lambda(h(p)+h(q)-K)$
.
Then$G_{p}=1+\lambda h’(p)=0,$ $G_{q}=1+\lambda h’(q)=0$
.
Because $h^{u}(x)>0$
,
we get $p=q$.
So
$p+q$ attains its extremum at $p=q$.
Wecan show that when$p=q,$ $p+q$ is minimumunder the restriction (A2).
Therefore
19
Hence it follows that
$p=q \geq\frac{p’+q’}{2}$
Thus we get
(A.3) $p+q\geq p’+q’$
.
which proves Lemma
3.5.
B Proof of Lemma 3.6
To prove Lemma 3.6, we must check the following 12 cases:
1) $c_{1n}<l$
,
(1) $S_{2}$ crosses
$i_{0_{n-}}^{n-}$,
(2) $R_{2}$ crosses $\iota_{0}$
$n-$ (3) no wave cross $\iota_{0}$
2) $c_{1n}\geq l$,
(1) $S_{2}$ and $S_{1}$ cross $i_{0}^{n-}$,
(2) $R_{2}$ and $S_{1}$ cross $i_{0}^{n-}$
,
(3) $S_{2}$ and $R_{1}$ cross $i_{0}^{n-}$
,
(4) $R_{2}$ and $R_{1}$ cross $i_{0}^{n-}$,
(5) $S_{1}$ crosses $i_{0_{n-}}^{n-}$, (6) $R_{1}$ crosses $\iota_{n^{0}-}$ , (7) $S_{2}$ crosses $\iota_{0}$ , (8) $R_{2}$ crosses $i_{0}^{n-},.n-$ (9) no wave cross $\iota_{0}$
Put $r_{+}^{n-1}=r^{l}(a_{1n-1}),$ $s_{+}^{n-1}=s^{l}(a_{1n-1}),$ $r_{-}^{n-1}=-s_{-}^{n-1}$
$=r^{l}((n-1)h+0,0)$, and $\delta_{n-1}=U^{l}(a_{1n-1})$
.
Put $r_{+}^{n-1’}=r^{l}((n-1)h+O, 2l)$ and $s_{+}^{n-1’}=s^{l}((n-1)h+O,2l)$
.
Put $A=(r_{-}^{n-1}, s_{-}^{n-1}),$ $B=(r_{+}^{n-}, s_{+}^{n-1})$ and $B’=(r_{+}^{n-1’}, s_{+}^{n-1’})$
.
Put $C=(r_{+}^{n-1}+Kh, s_{+}^{n-1}+Kh)$,
(resp $=(r_{+}^{n-1’}+\delta_{n-1}h,$$s_{+}^{n-1’}+\delta_{n-1}h,$$)$ ) if $c_{1n}<l$ (resp $c_{1n}\geq l$).
If$R_{2}$ crosses $i_{0}^{n+},$
$F(i_{0}^{n+_{n+}})=0\leq F(i_{0}^{n-})$, so that it is sufficient to consider the
cases when $S_{2}$ crosses $\iota_{0}$
.
Figure.2 1) $c_{1n}<l$
.
(1) $S_{2}$ crosses $i_{0}^{n-}$ (Figure 2). Denote by I (resp II) the halfspace
$\{(r,s)|r+s<0\}$ (resp $\{(r,s)|r+s\geq 0\}$
.
)i) $C\in I$
.
In this case $S_{2}$ crosses $i_{0}^{n+}$
.
Denote by V(PQ) the absolute value of thetotal variation of$r$ and $s$ by the line segment PQ. From Figure.3, $F(i_{0}^{n+})=V(A’C)\leq V(A’C’)=V(AB)=F(i_{0}^{n-})$
.
21
Figure.3 ii) $C\in II$
.
In this case $R_{2}$ crosses $i_{0}^{n+}$
.
Then(B.1) $F(i_{0}^{n-})\geq F(i_{0}^{n+})=0$
.
(2) $R_{2}$ crosses $i_{0}^{n-}$
In this case $B\in II$ so that $R_{2}$ crosses $i_{0}^{n+}$
.
Then(B.2) $F(i_{0}^{n-})=F(i_{0}^{n+})=0$
.
(3) no wave crosses $i_{0}^{n-}$
In this case $(r_{+}^{n-1}, s_{+}^{n-1})$ is on the line
$r+s=0$
.
Hence $C\in II$.
It is2) $c_{1n}\geq l$
.
(1) $S_{2}$ and $S_{1}$ cross $i_{0}^{n-}$ (Figure.4)
23
Figure.5 i) $C\in I$
.
From Figure.5,
$F(i_{0}^{n+})=V(A’C)\leq V(A’C’)=V(A^{u}B’)=V(AB’)=F(i_{0}^{n-})$
.
ii) $C\in II$ implies that $R_{2}$
crosses
$i_{0}^{n+}$.
So
we get (B2).(2) $R_{2}$ and $S_{1}$ cross $i_{0}^{n-}$ Figure.6 i) $C\in I$
.
From Figure.6, $F(i_{0}^{n+})=V(A’C)\leq V(A’D)=V(A^{u}E)=V(A^{u}B^{u})$ $\leq V(BB’’)=V(BB’)=F(i_{0}^{n+})$ii) $C\in II$
.
25
(3) $S_{2}$ and $R_{1}$ cross $i_{0}^{n-}$
Figure.7
Put $G=$ $(r_{+}^{n-} +\delta_{n-1}h, s_{+}^{n-1}+\delta_{n-1}h)$ and $II=(r^{l}(a_{1n}), s^{l}(a_{1n}))$
.
Then $H$ is on the line $CG$.
i) $H\in I$.
From Figure.7,
$F(i_{0}^{n+})=V(A’H)\leq V(A^{u}G)\leq V(AB)=F(i_{0}^{n-})$
.
ii) $H\in II$, so $R_{2}$ crosses $i_{0}^{n+}$, and we get
$(B2)n-$ (4) $R_{2}$ and $R_{1}$ cross $\iota_{0}$
(5) $S_{1}$ crosses $i_{0}^{n-}$ Figure.8 i) $C\in I$
.
From Figure.8, $F(i_{0}^{n+})=V(A’C)=V(AE)=V(AD)$ $\leq V(AB’)=F(i_{0}^{n-})$ Thus we get (B1). ii) $C\in II$.
27
(6) $R_{1}$ crosses $i_{0}^{n-}$
In this case, it is obvious that $F(i_{0}^{n+})=0$
.
IIence we get (B3).Cases (7), (8) and (9) are almost the same as cases (1), (2) and (3) in 1).
Thus, we obtain Lemma
3.6.
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